立志用最少的代碼做最高效的表達

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PAT甲級最優題解——>傳送門

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Calculate a+b and output the sum in standard format – that is, the digits must be separated into groups of three by commas (unless there are less than four digits).

Input Specification:
Each input file contains one test case. Each case contains a pair of integers a and b where ?10^?6≤a,b≤10^?6. The numbers are separated by a space.

Output Specification:
For each test case, you should output the sum of a and b in one line. The sum must be written in the standard format.

Sample Input:
-1000000 9
Sample Output:
-999,991

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技巧：

直接判斷后輸出負號，字符串存儲絕對值，方便統一計算。

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代碼一：正序處理

利用取余確定循環初值，然后再按3的倍數輸出字符串和逗號

#include<bits/stdc++.h> using namespace std; int main( ) {int a, b; cin >> a >> b;if(a + b < 0) putchar('-');string s = to_string(abs(a+b));int i = s.size()%3 == 0 ? 3 : s.size()%3;cout << s.substr(0, i);for(; i < s.size(); i += 3) cout << ',' << s.substr(i, 3);return 0; }

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代碼二：倒序處理

字符串倒置，按3的倍數輸出逗號和字符串即可。

#include<bits/stdc++.h> using namespace std; int main() {int a, b; cin >> a >> b;if(a + b < 0) cout << '-';string s = to_string(abs(a+b));vector<string>v;for(int i = s.size(); i > 0; i -= 3) v.push_back(i>=3 ? s.substr(i-3, 3) : s.substr(0, i));for(int i = v.size()-1; i >= 0; --i)printf("%s%s", v[i].c_str(), i>0?",":""); }

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耗時

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